Problem: ${\sqrt[3]{343} = \text{?}}$
Solution: $\sqrt[3]{343}$ is the number that, when multiplied by itself three times, equals $343$ If you can't think of that number, you can break down $343$ into its prime factorization and look for equal groups of numbers. So the prime factorization of $343$ is $7\times 7\times 7$ We're looking for $\sqrt[3]{343}$ , so we want to split the prime factors into three identical groups. We only have three prime factors, and we want to split them into three groups, so this is easy. $343 = 7\times 7\times 7$ , so $7^3 = 343$ So $\sqrt[3]{343}$ is $7$.